Question: Caterina scans animals brought to the shelter to check for microchips that will help locate their owners. There is a $0.05$ probability that a stray dog brought to the shelter will have a microchip. Let $D$ be the number of stray dogs Caterina scans until she finds one with a microchip. Assume the probability of each dog having a microchip is independent. Find the probability that the $4^{\text{th}}$ dog Caterina scans will be the first to have a microchip. You may round your answer to the nearest hundredth. $P(D=4)=$
Answer: Without a fancy calculator For each dog: $P({\text{microchip}})=0.05$ $P(\text{none}})=0.95$ If the $4^{\text{th}}$ dog Caterina scans will be the first to have a microchip, her sequence of results from the scans must be "none, none, none, microchip." $\begin{aligned} P(D=4)&=P(\text{NNN}}{\text{M}}) \\\\ &=(0.95})(0.95})(0.95})({0.05}) \\\\ &=(0.95)^3(0.05) \\\\ &=0.04286875 \end{aligned}$ $P(D=4) = 0.04286875 \approx 0.04$